EMA4314_HW7_solutions

EMA4314_HW7_solutions - EMA 4314 HW7 solutions 20.22 (a) 1...

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EMA 4314 HW7 solutions 20.22 (a) D = 1 3 λ c ' = kT σ P c ' = 8 RT π M eq. 20.22, 20.13, 20.7 a. D = ( 1 3 )( (1.38 × 10 23 J K )(298 K ) (3.6 × 10 19 m 2 )(1 Pa ) (8*8.3145 J K * mol )(298 K ) * 39.948 × 10 3 kg mol = 1.51 m 2 s This value differs from the book by a factor of 2/3, which is explained on page 775. b. D = 1.51x10 -5 m 2 s -1 c. D = 1.51x10 -7 m 2 s -1 J = D N dP dx N = P RT dP dx = 0.1 atm cm = 1.013 × 10 6 Pa m a. J = 617 mol/m 2 *s b. J = 617 x 10 -3 mol/m 2 *s c. J = 617 x 10 -7 mol/m 2 *s 20.30 (a) < x 2 > 1/ 2 = Dt t = < x 2 > 2 D t = (5.0 × 10 3 m ) 2 2(3.17
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This note was uploaded on 11/16/2011 for the course EMA 4314 taught by Professor Phillpot during the Fall '10 term at University of Florida.

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EMA4314_HW7_solutions - EMA 4314 HW7 solutions 20.22 (a) 1...

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